1
$\begingroup$

Can you give any hints how to solve the integral $$ \int\limits_{0}^{\infty} {\exp \left( \frac{-w (z-u)^2}{2u^2 z} \right) dz} $$ for $u>0, w>0$, or does no close form exist? Substitution seems to be difficult…

$\endgroup$
  • $\begingroup$ I would suggest expanding the square: $\frac{-w(z- u)^2}{2\mu^2z}= \frac{-wz^2+ 2wuz- wu^2}{2\mu z}= \frac{-w}{2\mu}z+ \frac{wu}{\mu}- \frac{wu^2}{2\mu}\frac{1}{z}$. The "1/z" is the hard part. $\endgroup$ – user247327 Jun 7 '17 at 19:45
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}\exp\pars{-\,{w\bracks{z - u}^{2} \over 2u^{2}z}}\,\dd z = \int_{0}^{\infty}\exp\pars{-\,{w \over 2u^{2}}\, \bracks{\root{z} - {u \over \root{z}}}^{2}}\,\dd z \\[5mm] \stackrel{\root{z}\ =\ \root{u}\expo{\theta}}{=}\,\,\,& \int_{-\infty}^{\infty}\exp\pars{-\,{2w \over u}\, \sinh^{2}\pars{\theta}}\pars{2u\expo{2\theta}}\,\dd\theta \\[5mm] = &\ u\int_{-\infty}^{\infty}\exp\pars{-\,{2w \over u}\, {\cosh\pars{2\theta} - 1 \over 2}}\bracks{\sinh\pars{2\theta} + \cosh\pars{2\theta}}\,2\,\dd\theta \\[5mm] = &\ 2u\expo{w/u}\int_{0}^{\infty}\exp\pars{-\,{w \over u}\, \cosh\pars{\theta}}\cosh\pars{\theta}\,\dd\theta \\[5mm] = &\ \left.-2u\expo{w/u}\partiald{}{z}\int_{0}^{\infty} \exp\pars{-\,z\cosh\pars{\theta}}\,\dd\theta\,\right\vert_{\ z\ =\ w/u} \\[5mm] = &\ -2u\expo{w/u}\,\underbrace{\mrm{K}_{0}'\pars{w \over \mu}} _{\ds{-\,\mrm{K}_{1}\pars{w/u}}}\,,\qquad\qquad\qquad\qquad\qquad\qquad \verts{\mrm{arg}\pars{w \over u}} < {\pi \over 2} \end{align}

$\ds{\,\mrm{K}_{\nu}}$ is a Modified Bessel Function. See A & S Table.


$$ \bbx{\int_{0}^{\infty}\exp\pars{-\,{w\bracks{z - u}^{2} \over 2u^{2}z}}\,\dd z = 2u\expo{w/u}\,\mrm{K}_{1}\pars{w \over u}} \qquad\qquad \verts{\mrm{arg}\pars{w \over u}} < {\pi \over 2} $$

$\endgroup$
  • $\begingroup$ @jana1802 Fixed. Thanks. It's funny yesterday it was right but last night I 'recheck' everything and made the mistake you pointed out. $\endgroup$ – Felix Marin Jun 9 '17 at 6:56
1
$\begingroup$

By expanding the square and performing a substitution we have that the given integral depends on a modified Bessel function of the second kind, since

$$ \forall \alpha>0,\qquad \int_{0}^{+\infty}\exp\left(-\alpha\left(w+\frac{1}{w}\right)\right)\,dw = 2\cdot K_1(2\alpha).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.